Re: equivalence

Now to show 1>0 ,YOU MUST SHOW : 1.c>0.c & c>0,so that you can use M.Ponens and get 1>0

Deveno did not show that ,he simply assumed it.

He assumed : c>0

But c>0 => 1.c>0.c ,because 1.c=c and 0.c =0

Hence by M.Ponens we have : 1.c>0.c.

Thus , 1.c>0.c & c>0

But by using (3) we have : 1.c>0.c &c>0 => 1>0.

And by using M.Ponens again we have : 1>0.

So far we have ASSUMED , c>0 and ended up with 1>0 ..

But ,to be exact ,we also have assumed : ac>bc &c>0 => a>b

So we have proved : [(ac>bc& c>0) =>a>b]& c>0 => 1>0 and not 1>0.

IF you assume ,p and end with q ,you have proved : p=>q and not q

We want to show (3) --> (1). Therefore, we assume (3) to be true, and derive (1) conditionally.

Now (3) itself is a statement of the form: (p&q)-->r. Thus in taking (3) to be true, we are taking it to be of the form: T-->T.

I have NOT shown that , in fact, 1 > 0. I have shown that if (3) is true, for any $a,b \in \Bbb R$, and $c \in \Bbb R^+$, then in particular it is true for some particular pair $a,b$, namely $a = 1,b = 0$.

If you want to QUIBBLE, the only "hole" is that the set $\Bbb R^+$ may be empty. This is actually somewhat problemmatic, as one has to define exactly what is MEANT by $x > y$, for two real numbers $x,y$.

The standard way of doing this, is to use the density of the rational numbers in the reals, and induce the ordering on $\Bbb R$ from that in $\Bbb Q$. That is, we say, for two real numbers $x,y$, that $x < y$, if for every rational number $p < x$, and every rational number $y < q$, we have $p < q$. To an extent, this "begs the question": what does it mean, in the rationals, for $p < q$ to be true? (Note that this specifically addresses the fact that the ordering of the rationals in Archimedean, and the the supremum of a set of rational numbers is a real number (the "least upper bound" property of the reals)).

Here, we resort to the ordering on the integers: we say that $p = \dfrac{a}{b} < q = \dfrac{c}{d}$, if $ad < bc$. Again, we must ask: how do we define the ordering on the integers?

We say that for two integers: $m > n$, if $m - n = m + (-n)$ is a (non-zero) natural number. Alternatively, writing an integer as a an equivalence of pairs of natural numbers $[(a,b)]$ ($a$ is the "positive part" and $b$ is the "negative part"), we say that $[(a,b)]$ is positive (> 0 = [(0,0)]) if $a > b$, as natural numbers. So, again, we have "pushed the problem" to yet another ordered set, the natural numbers.

For two natural numbers, $k,k'$ we say that $k > k'$ if there exists a natural number $n$ such that $k$ is the $n$-th successor of $k'$.

This means that the integer $k$, that is the equivalence class of $(k,0)$, for a non-zero natural number $k$ is positive (greater than 0), whence it follows that the rational number $\dfrac{k}{1}$ is likewise positive (greater than $0 = \dfrac{0}{1}$), and thus that real number $k$ is also greater than 0. This shows that we have (at least) an infinite choice of real numbers $c > 0$ we might choose from (any real form of a natural number).

Of course, 1 is one of those numbers, so it appears that we have inadvertently used "circular reasoning" (used (1) to assert the existence of $c$).

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So, can we justify that there exists at least ONE positive real number (without appealing to $1$)? I claim we can, if we can use THESE axioms (in addition to the field axioms):

A1) $a > 0,b > 0 \implies ab > 0$, for any $a,b \in \Bbb R$.

A2) $a > 0$ or $-a > 0$, or $a = 0$, and exactly one of these holds.

Claim: if $a \neq 0$, then $a^2 > 0$.

Since $\Bbb R$ is a field, if $a^2 = 0$, then either $a = 0$, or...$a = 0$ (fields are integral domains, and posses no zero divisors). Since $a \neq 0$, it follws $a^2 \neq 0$, and thus by (A2) either $a^2 > 0$, or $-a^2 > 0$.

Now, also by (A2), since $a \neq 0$, we have either $a > 0$ or $a < 0$. If $a > 0$, then by (A1), we have $a^2 > 0$.

By the way, my point was, $\displaystyle 1>0$ is a basic axiom of integers and continues to hold for the integers embedded within the real numbers. Hence, any implication of the form $\displaystyle P \Rightarrow 1>0$ is true regardless of the statement $\displaystyle P$.